Problem: Let $f(x)=2^{(5x-3x^2)}$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $(5-6x)\cdot 2^{(5x-3x^2)}$ (Choice B) B $\ln(2)\cdot 2^{(5x-3x^2)}$ (Choice C) C $\ln(2)\cdot 2^{(5x-3x^2)}\cdot (5-6x)$ (Choice D) D $\ln(5-6x)\cdot 2^{(5x-3x^2) }\cdot 2$
Answer: $f$ is an exponential function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=5x-3x^2$, then $f(x)=2^{u(x)}$. $f'(x)$ can be found using the following identity: $\dfrac{d}{dx}\left[2^{u(x)}\right]=\ln(2)\cdot 2^{u(x)}\cdot u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}2^{(5x-3x^2)} \\\\ &=\dfrac{d}{dx}2^{u(x)}&&\gray{\text{Let }u(x)=5x-3x^2} \\\\ &=\ln(2)\cdot 2^{u(x)}\cdot u'(x) \\\\ &=\ln(2)\cdot 2^{(5x-3x^2)}\cdot (5-6x)&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ In conclusion, $f'(x)=\ln(2)\cdot 2^{(5x-3x^2)}\cdot (5-6x)$.